计算:cos(70°+α)sin(170°-α)-sin(70°+α)cos(10°+α)
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 20:14:38
计算:cos(70°+α)sin(170°-α)-sin(70°+α)cos(10°+α)
sin x = sin (180° - x)
所以,
sin (170° - a) = sin [180° - (170° - a)]
= sin (10° + a)
根据方程式,
sin(u ±v) = sin u cos v ± cos u sin v
sin (10° + a) cos (70° + a) - cos (10° + a) sin (70° + a)
= sin [(10° + a) - (70° + a)]
= sin - 60°
= - sin 60°
= -(√3)/2
= -0.866
纯手打,蒙采纳,谢谢~~
所以,
sin (170° - a) = sin [180° - (170° - a)]
= sin (10° + a)
根据方程式,
sin(u ±v) = sin u cos v ± cos u sin v
sin (10° + a) cos (70° + a) - cos (10° + a) sin (70° + a)
= sin [(10° + a) - (70° + a)]
= sin - 60°
= - sin 60°
= -(√3)/2
= -0.866
纯手打,蒙采纳,谢谢~~
化简:(2)cos(70°+α)sin(170°+α)-sin(70°+α)cos(10°+α)
Cos(180°+α)*sin(α+360°)/sin(-α-180°)*cos(-180°-α)
sinα + cosα
sinα+cosβ=sin225° sinα-cosβ=cos225°求cos(α-β)及cos(α+β)
化简cos^2(-α)-[tan(360°+α)/sin(-α)]
化简,cos²(-α)-tan(360°+α)/sin(-α)
-cos(α-45°)=sin(α+45°)对吗?
已知sinαcosα= ,且0°
已知sinαcosα=0.125,且0°
化简cos(540°-α)sin(α-360°)/sin(-α+180°)cos(180°+α)
化简cos(α+30°)cos(α-30°)-sin(α+30°)sin(α+150°),
.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3